A high-speed passenger train travels from Station X to Station Y at a constant speed of 60… — GMAT

A high-speed passenger train travels from Station X to Station Y at a constant speed of $60$ miles per hour. On the return trip along the exact same track, due to maintenance, the train travels at a constant speed of $40$ miles per hour. What is the average speed of the train, in miles per hour, for the entire round trip?

A
$45$
$48$ Answer
C
$50$
D
$52$
E
$54$

Answer

48

miles per hour

Average speed is defined as total distance divided by total time. Setting distance as

d

, the outbound travel time is

\frac{d}{60}

and the return time is

\frac{d}{40}

. The total distance is

2d

and total time is

\frac{5d}{120} = \frac{d}{24}

. Dividing total distance

2d

by total time

\frac{d}{24}

gives

48

miles per hour.

Step-by-Step Solution

Define variables for distance and time for each leg of the journey.

Let the distance between Station X and Station Y be

d

miles. Outbound time

t_1 = \frac{d}{60}

hours, and return time

t_2 = \frac{d}{40}

hours.

Average speed is defined as total distance divided by total time, not the simple average of speeds.

Calculate total distance and total time.

Total distance

= d + d = 2d

. Total time

= t_1 + t_2 = \frac{d}{60} + \frac{d}{40} = \frac{2d + 3d}{120} = \frac{5d}{120} = \frac{d}{24}

hours.

Summing individual distances and times provides the components needed for the average speed formula.

Divide total distance by total time to find the average speed.

Average speed

= \frac{2d}{\frac{d}{24}} = 2d \times \frac{24}{d} = 48

miles per hour.

Executing the division yields the overall rate for the trip.

Key Concept

Average Speed for Multi-Leg Trips

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